What is the right notion of transcendence degree for the fraction field of an affinoid algebra

Question

Let $A$ be a finitely generated $k$-algebra with fraction field $K$. Then the Krull dimension of $A$ is equal to the transcendence degree of $K$ over $k$.

I would be very interested in any related result where $A$ is an affinoid algebra. I suppose then $K$ needs not be of finite transcendence degree. Is there any reasonable analogue of a finite transcendence basis that somehow takes the norm of $A$ into account?

Many thanks for answering!

Answer 1

The fact that you are using $K=\mathrm{Frac}(A)$ is misleading. The fraction field is only relevant because $K$ is the residue field of a point of $\mathrm{Spec}(A)$ — it is $k(\eta)$ for $\eta$ the generic point of $\mathrm{Spec}(A)$.

So, perhaps a more true-to-geometry way to write your formula is the following. Suppose that $A$ is a finite type $k$-algebra. Then,

$$\displaystyle \dim(A)=\sup _{x\in \mathrm{Spec}(A)} \mathrm{tr.deg}(k(x)/k).$$

What this also highlights is also that you have to use $\mathrm{Spec}(A)$ and not $\mathrm{MaxSpec}(A)$ — you really need ‘non-classical points’ (i.e., not $\overline{k}$-points) to get the correct answer.

With these comments in mind, then a similar formula holds in rigid geometry. If $k$ is a non-archimedean field, and $A$ is an affinoid $k$-algebra (i.e., a topological $k$-algebra which is topologically of finite type), then

$$\dim(A)=\sup_{x\in\mathrm{Spa}(A)}\mathrm{trc.deg}(k(x)/k).$$

Let me explain the notation here:

• $\mathrm{Spa}(A)=\mathrm{Spa}(A,A^\circ)$ is the adic space associated to $A$ — this shows up, and not something like the Tate spectrum $\mathrm{Sp}(A)$, for the same reason as before: you need to consider all, in particular non-classical, points to get the right formula,

• $\mathrm{trc.deg}(k(x)/k)$ is a topological variant of transcendence degree, which is defined as the minimum of $\mathrm{tr.deg}(L/k)$ where $L$ runs over subextensions of $k(x)/k)$ which are ‘algebraically dense’ — you can obtain $L$ from $k$ by applying the operation $\widetilde{(-)}$ (i.e., the topological closure of the algebraic closure) finitely many times.

Remark: Because $k(x)=k(y)$ if $x$ and $y$ are related by specializations/generalizations, you can also replace $\mathrm{Spa}(A)$ by its Berkovich space $\mathcal{M}(A)$.